Description

Given a binary tree, find the lowest common ancestor (LCA) of two given nodes in the tree.

According to the definition of LCA on Wikipedia: “The lowest common ancestor is defined between two nodes p and q as the lowest node in T that has both p and q as descendants (where we allow a node to be a descendant of itself).”

Given the following binary tree: root = [3,5,1,6,2,0,8,null,null,7,4]

Example 1:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 1
Output: 3
Explanation: The LCA of nodes 5 and 1 is 3.

Example 2:

Input: root = [3,5,1,6,2,0,8,null,null,7,4], p = 5, q = 4
Output: 5
Explanation: The LCA of nodes 5 and 4 is 5, since a node can be a descendant of itself according to the LCA definition.

Note:

  • All of the nodes’ values will be unique.
  • p and q are different and both values will exist in the binary tree.

Solutions

1. Recursion

# Time: O(logn)
# Space: O(logn)
# Definition for a binary tree node.
# class TreeNode:
#     def __init__(self, x):
#         self.val = x
#         self.left = None
#         self.right = None

class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        if not root or root == p or root == q:
            return root
        
        left = self.lowestCommonAncestor(root.left, p, q)
        right = self.lowestCommonAncestor(root.right, p, q)
        return root if left and right else left or right

# 31/31 cases passed (68 ms)
# Your runtime beats 88.17 % of python3 submissions
# Your memory usage beats 91.67 % of python3 submissions (23.1 MB)

2. Iterative

# Time: O(logn)
# Space: O(logn)
class Solution:
    def lowestCommonAncestor(self, root: 'TreeNode', p: 'TreeNode', q: 'TreeNode') -> 'TreeNode':
        stack = [root]
        parent = {root: None}
        while p not in parent or q not in parent:
            node = stack.pop()
            if node.left:
                parent[node.left] = node
                stack.append(node.left)
            if node.right:
                parent[node.right] = node
                stack.append(node.right)
        ancestors = set()
        while p:
            ancestors.add(p)
            p = parent[p]
        while q not in ancestors:
            q = parent[q]
        return q

# 31/31 cases passed (68 ms)
# Your runtime beats 88.17 % of python3 submissions
# Your memory usage beats 100 % of python3 submissions (16.5 MB)

References

  1. 236. Lowest Common Ancestor of a Binary Tree